Question

Out of 7 consonants and 4 vowels, words are to be formed each having only 3 consonants and 2 vowels. The number of such words formed is :
(a)210
(b)25200
(c)2520
(d)302400

Answer 1

Hint: Use the formula: Number of words formed = Number of ways of selecting 3 consonants $\times $ number of ways of selecting 2 vowels $\times $ total permutations of 5 alphabets .

Complete step by step answer:
Number of ways of selecting 3 consonants is ${}^{7}{{C}_{3}}=35$, number of ways of selecting 2 vowels is ${}^{4}{{C}_{2}}=6$, and total permutations of 5 alphabets is $5!=120$.
We are given that out of 7 consonants and 4 vowels, words are to be formed each having only 3 consonants and 2 vowels.
We need to find the number of such words that can be formed.
We will use the principle of combinations to solve this question.
We know that the number of ways to choose r things from a total of n things is ${}^{n}{{C}_{r}}$
${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$
In this question, we will need 3 consonants out of the 7 consonants and 2 vowels out of the 4 vowels.
First, we will find the number of ways to select 3 out of 7 consonants. This is given by:
${}^{7}{{C}_{3}}=\dfrac{7!}{3!4!}$
${}^{7}{{C}_{3}}=35$ …(1)
Now, we will find the number of ways to select 2 out of 4 consonants. This is given by:
${}^{4}{{C}_{2}}=\dfrac{4!}{2!2!}$
${}^{4}{{C}_{2}}=6$ …(2)
So, now we have all the letters which are required.
We now have to arrange these letters to form different words by making different permutations using these letters.
We know that the number of permutations of n alphabets is equal to $n!$.
We have 3 consonants and 2 vowels which makes it 5 alphabets.
So, the number of permutations using 5 alphabets is equal to $5!=120$ …(3)
We will now find the total number of required words formed by multiplying (1), (2), and (3)
Number of words formed = Number of ways of selecting 3 consonants $\times $ number of ways of selecting 2 vowels $\times $ total permutations of 5 alphabets
Number of words formed$={}^{7}{{C}_{3}}\times {}^{4}{{C}_{2}}\times 5!$
                   $=35\times 6\times 120=25200$
So, the total number of words formed are 25200

Hence, option (b) is correct.

Note: In this question, it is very important to know the following formulae: the number of ways to choose r things from a total of n things is ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$, and the number of permutations of n alphabets is equal to $n!$.